Simplify and expand the following expression: $ \dfrac{2r}{3r + 5}+\dfrac{3r}{r - 1} $
Explanation: In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(3r + 5)(r - 1)$ Multiply the first term by $\dfrac{r - 1}{r - 1}$ $ \begin{align*} \dfrac{2r}{3r + 5} \times \dfrac{r - 1}{r - 1} & = \dfrac{(2r)(r - 1)}{(3r + 5)(r - 1)} \\ & = \dfrac{2r^2 - 2r}{(3r + 5)(r - 1)}\end{align*} $ Multiply the second term by $\dfrac{3r + 5}{3r + 5}$ $ \begin{align*} \dfrac{3r}{r - 1} \times \dfrac{3r + 5}{3r + 5} & = \dfrac{(3r)(3r + 5)}{(r - 1)(3r + 5)} \\ & = \dfrac{9r^2 + 15r}{(r - 1)(3r + 5)}\end{align*} $ Now we have: $ = \dfrac{2r^2 - 2r}{(3r + 5)(r - 1)} + \dfrac{9r^2 + 15r}{(r - 1)(3r + 5)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{2r^2 - 2r + 9r^2 + 15r}{(3r + 5)(r - 1)} $ $ = \dfrac{11r^2 + 13r}{(3r + 5)(r - 1)}$ Expand the denominator: $ = \dfrac{11r^2 + 13r}{3r^2 + 2r - 5}$